Question

F(x)=x^2+√x производной

Answer 1

$f(x)=x^2+ \sqrt{x} \\ f'(x)=2x+ \frac{1}{2 \sqrt{x} }$

Answer 2

To find the derivative of the given function $F(x) = x^2 + \sqrt{x}$ with respect to $x$, we can use the rules of differentiation. The general power rule and the chain rule will be applied.

The power rule states that if we have a term of the form $x^n$, the derivative with respect to $x$ is $nx^{(n-1)}$. And for the square root term, the derivative is $\frac{1}{2\sqrt{x}}$.

Now, let's find the derivative of $F(x)$:

$F(x) = x^2 + \sqrt{x}$

Applying the power rule to the $x^2$ term, we get:

$\frac{d}{dx}(x^2) = 2x^{(2-1)} = 2x$

And applying the chain rule to the $\sqrt{x}$ term, we get:

$\frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}} \cdot \frac{d}{dx}(x) = \frac{1}{2\sqrt{x}} \cdot 1 = \frac{1}{2\sqrt{x}}$

Now, adding both derivatives together, we get the derivative of the entire function:

$F'(x) = \frac{d}{dx}(x^2) + \frac{d}{dx}(\sqrt{x}) = 2x + \frac{1}{2\sqrt{x}}$

So, the derivative of $F(x) = x^2 + \sqrt{x}$ with respect to $x$ is $F'(x) = 2x + \frac{1}{2\sqrt{x}}$.